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A circular plate of radius R/2 is cut from one edge of a thin circular plate of radius R. The moment of inertia of the remaining portion about an axis through O perpendicular to plane of the plate is (M = mass of remaining circular plate.)

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$\frac{5}{7} {MR}^{2}$

$\frac{13}{24} {MR}^{2}$

$\frac{7}{12} {MR}^{2}$

$\frac{13}{32} {MR}^{2}$

answer is A.

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Detailed Solution

Let m0 be the total mass of plate. Mass of removed plate $m_{1} = \frac{m_{0}}{4}$

Mass of remaining plate is $m = \frac{{3m}_{0}}{4}$

$I_{total} {=I}_{removel} {+ I}_{remaining} \Rightarrow I_{remaining} {= I}_{total} {- I}_{removed}$

$m_{0} \frac{R^{2}}{2} - (\frac{\frac{m_{0}}{4} {(\frac{R}{2})}^{2}}{2} + \frac{m_{0}}{4} {(\frac{R}{2})}^{2}) = \frac{13}{24} {mR}^{2}$

When $m_{0} = \frac{4m}{3}$

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